If #f(x) =sec^3(x/2) # and #g(x) = sqrt(2x-1 #, what is #f'(g(x)) #?

2 Answers
Jan 21, 2016

Answer:

#f'(g(x))=3/2sec^3(sqrt(2x-1)/2)tan(sqrt(2x-1)/2)#

Explanation:

First, we should just find #f'(x)# so that we can then plug #g(x)# into it.

To find #f'(x)#, the primary and overriding issue is the third power. According to the chain rule, #d/dx[u^3]=3u^2*u'#, hence

#f'(x)=3sec^2(x/2)*d/dx[sec(x/2)]#

Now, to differentiate the secant function, recall that (through the chain rule) #d/dx[sec(u)]=sec(u)tan(u)*u'#, hence

#f'(x)=3sec^2(x/2)*sec(x/2)tan(x/2)*d/dx[x/2]#

Note that #d/dx[x/2]=d/dx[1/2x]=1/2#. Also, recognize that the secant functions can be multiplied with one another, yielding a simplified #f'(x)#:

#f'(x)=3/2sec^3(x/2)tan(x/2)#

To find #f'(g(x))#, simply plug in #sqrt(2x-1)# wherever there's an #x# in #f'(x)#.

#f'(g(x))=3/2sec^3(sqrt(2x-1)/2)tan(sqrt(2x-1)/2)#

Jan 21, 2016

Answer:

#f'(g(x))=3/(2(sqrt(2x-1)))sec^3(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)#

Explanation:

If the question meaning is #f'(g(x))# as #(f@g)'# then:

find #f@g=f(g(x))#

#y=f(g(x))=[sec(sqrt(2x-1)/2)]^3#

#g(x)# is a composition too, therefore you can think:

#y=u^3#

#u=sec(v)#

#v=sqrt(z)/2#

#z=(2x-1)#

the Chain Rule tells

#dy/dx=f'(g(x))=(dy)/(du)(du)/(dv)(dv)/(dz)(dz)/(dx)#

#dy/(du)=3u^2=3sec^2(v)=3sec^2(sqrt(z)/2)=3sec^2(sqrt(2x-1)/2)#

#(du)/(dv)=sec'(v)=tan(v)*sec(v)=tan(sqrt(z)/2)*sec(sqrt(z)/2)=#
#=tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)#

#(dv)/(dz)=1/2*(1/(2sqrt(z)))=1/(4sqrt(2x-1))#

#(dz)/(dx)=2*1+0#

Thus:

#y'=3sec^2(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)*1/(cancel(4)^2sqrt(2x-1))*cancel(2)=#
#=3/(2(sqrt(2x-1)))sec^2(sqrt(2x-1)/2)*tan(sqrt(2x-1)/2)*sec(sqrt(2x-1)/2)#

Alternatively, once you get practice, more simply:

Given:

#h(x)=f(g(x))#

#h'(x)=f'(g(x))*g'(x)#

if #g(x)=g(i(x)) =>g'(x)=g'(i(x))*i'(x)#

thus you can apply the rule ricorsively.